Can Operators A and B Always Be Applied to a General State in Tensor Space?

[Kreizhn]

Hey,

My brain seems to have shut down. Let's say I'm working in the space $H_a \otimes H_b$ and I have an operator A and B on each of these spaces respectively. Furthermore, consider a general state in the tensor space $|\psi \rangle = | x_1 \rangle |y_1\rangle + |x_2 \rangle |y_2 \rangle$ (ignoring normalization). Now in my head, I keep on thinking that if I wanted to apply the operator $A \otimes B$ to $| \psi \rangle$ I couldn't simply go

$$A \otimes B | \psi \rangle = A| x_1 \rangle B|y_1\rangle + A|x_2 \rangle B|y_2 \rangle$$

However, I was playing around with some values today (assuming finite dimensions), and from what I tried it always seemed to work. Can we do this in general, or did I just get lucky in all my trials?

 

[javierR]

It's fine. First, AxB is defined so that (AxB)(|y> x |z>)=(A|y> x B|z>), and second, the operator (AxB) is linear on the tensor space...call $$(A\times B)=O$$ and $$|y> \times |z>=\Psi$$. Then $$O (\Psi_1+\Psi_2)=(O\Psi_{1}+O\Psi_{2})$$. Then use the first property for each of the terms. (the "x" is product)

 

[Entropia]

Hey there! It's totally understandable that your brain may feel a bit overwhelmed when thinking about operators on tensor states. Let's break it down and see if we can make it a bit clearer for you.

First, it's important to note that the tensor product of two Hilbert spaces, denoted as H_a \otimes H_b, is a new Hilbert space formed by the combination of the two original spaces. This means that any operators acting on this new space must also be defined on the tensor product space.

Now, let's consider your general state |\psi \rangle = | x_1 \rangle |y_1\rangle + |x_2 \rangle |y_2 \rangle. This state is a combination of two basis states, one from H_a and one from H_b. So, when we apply the operator A \otimes B to this state, we are essentially applying A to the first basis state and B to the second basis state.

In other words, A \otimes B |\psi \rangle = A| x_1 \rangle B|y_1\rangle + A|x_2 \rangle B|y_2 \rangle. This is because A and B are defined on the tensor product space.

Now, whether this works for all cases or not depends on the specific operators A and B and the state |\psi \rangle. In general, if A and B are defined on the tensor product space, then this will work. However, if they are defined on different spaces, then this may not be the case.

I hope this helps clarify things a bit for you. Remember, operators on tensor states can be tricky, so don't be too hard on yourself if it takes some time to fully understand it. Keep exploring and playing with different values and states, and you'll get the hang of it in no time!